hdu5349__MZL's simple problem

MZL's simple problem

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1489    Accepted Submission(s): 602

Problem Description

A simple problem
Problem Description
You have a multiple set,and now there are three kinds of operations:
1 x : add number x to set
2 : delete the minimum number (if the set is empty now,then ignore it)
3 : query the maximum number (if the set is empty now,the answer is 0)

 

Input

The first line contains a number N (N≤106),representing the number of operations.
Next N line ,each line contains one or two numbers,describe one operation.
The number in this set is not greater than 109.

 

Output

For each operation 3,output a line representing the answer.

 

Sample Input

61 21 331 31 43

 

Sample Output

34

 

Source

2015 Multi-University Training Contest 5

 

题意：

当你输入1与x的时候，表示把x加入到集合中；

输入2时，表示删除集合中最小的数；

输入3时，表示输出最大的数。

这里博主偷懒选择了用STL里的multiset这个容器来操作；multiset和set操作基本一致，但multiset能存储相同的数据。代码如下：

#include<stdio.h>#include<iostream>#include<algorithm>#include<set>using namespace std;int main(){    int n,a,x;    multiset<int> s;    multiset<int>::iterator it;    scanf("%d",&n);    while(!s.empty()) {        s.erase(s.begin());    }    for(int i=0;i<n;i++) {        scanf("%d",&a);        if(a==1) {            scanf("%d",&x);            s.insert(x);        }        else if(a==2) {            if(s.size()) {                it=s.begin();                s.erase(it);            }        }        else if(a==3) {            if(s.size()) {                it=s.end();                it--;                printf("%d\n",*it);            }            else                printf("0\n");        }    }    return 0;}